elementary matrix operations as rank preserving operations

Let $M$ be a matrix over a division ring $D$ . An elementary operation on $M$ is any one of the eight operations below:

1.
exchanging two rows
2.
exchanging two columns
3.
adding one row to another
4.
adding one column to another
5.
right multiplying a non-zero scalar to a row
6.
left multiplying a non-zero scalar to a row
7.
right multiplying a non-zero scalar to a column
8.
left multiplying a non-zero scalar to a column

We want to determine the effects of these operations on the various ranks of $M$ . To facilitate this discussion, let $M=(a_{ij})$ be an $n\\times m$ matrix and $M^{\\prime}=(b_{ij})$ be the matrix after an application of one of the operations above to $M$ . In addition, let $v_{i}=(a_{i1},\\cdots,v_{im})$ be the $i$ -th row of $M$ , and $w_{i}=(b_{i1},\\cdots,b_{im})$ be the $i$ -th row of $M^{\\prime}$ . In other words,

M=\\begin{pmatrix}v_{1}\\\\\\vdots\\\\v_{n}\\end{pmatrix}=\\begin{pmatrix}a_{11}&\\cdots&a_{1m}\\\\\\vdots&\\ddots&\\vdots\\\\a_{n1}&\\cdots&a_{nm}\\end{pmatrix}\\xrightarrow[\\textrm{operation}]{\\textrm{%elementary}}\\begin{pmatrix}b_{11}&\\cdots&b_{1m}\\\\\\vdots&\\ddots&\\vdots\\\\b_{n1}&\\cdots&b_{nm}\\end{pmatrix}=\\begin{pmatrix}w_{1}\\\\\\vdots\\\\w_{n}\\end{pmatrix}=M^{\\prime}

Finally, let $d$ be the left row rank of $M$ .

Proposition 1.

Row and column exchanges preserve all ranks of $M$ .

Proof.

Clearly, exchanging two rows of $M$ do not change the subspace generated by the rows of $M$ , and therefore $d$ is preserved.

As exchanging rows do not affect $d$ , let us assume that rows have been exchanged so that the first $d$ rows of $M$ are left linearly independent.

Now, let $M^{\\prime}$ be obtained from $M$ by exchanging columns $i$ and $j$ . So $w_{1},\\ldots,w_{n}$ are vectors obtained respectively from $v_{1},\\ldots,v_{n}$ by exchanging the $i$ -th and $j$ -th coordinates. Suppose $r_{1}w_{1}+\\cdots+r_{d}w_{d}=0$ . Then we get an equation $r_{1}b_{1k}+\\cdots+r_{d}b_{dk}=0$ for $1\\leq k\\leq m$ . Rearranging these equations, we see that $r_{1}v_{1}+\\cdots+r_{d}v_{d}=0$ , which implies $r_{1}=\\cdots=r_{d}=0$ , showing that $w_{1},\\ldots,w_{d}$ are left linearly independent. This means that $d$ is preserved by column exchanges.

Preservation of other ranks of $M$ are similarly proved.∎

Proposition 2.

Additions of rows and columns preserve all ranks of $M$ .

Proof.

Let $M^{\\prime}$ be the matrix obtained from $M$ by replacing row $i$ by vector $v_{i}+v_{j}$ , and let $V^{\\prime}$ be the left vector space spanned by the rows of $M^{\\prime}$ . Since $v_{i}+v_{j}\\in V$ , we have $V^{\\prime}\\subseteq V$ . On other hand, $v_{i}=(v_{i}+v_{j})-v_{j}\\in V^{\\prime}$ , so $V\\subseteq V^{\\prime}$ , and hence $V=V^{\\prime}$ .

Next, let $w_{1},\\ldots,w_{n}$ be vectors obtained respectively from $v_{1},\\ldots,v_{n}$ such that the $i$ -th coordinate of $w_{k}$ is the sum of the $i$ -th coordinate of $v_{k}$ and the $j$ -th coordinate of $v_{k}$ , with all other coordinates remain the same. Again, by renumbering if necessary, let $v_{1},\\ldots,v_{d}$ be left linearly independent. Suppose $r_{1}w_{1}+\\cdots+r_{i}w_{i}+\\cdots+r_{d}w_{d}=0$ . A similar argument like in the previous proposition shows that $r_{1}v_{1}+\\cdots+(r_{i}+r_{j})v_{j}+r_{d}v_{d}=0$ , which implies $r_{1}=\\cdots=r_{i}+r_{j}=\\cdots r_{d}=0$ . Since $r_{i}=0$ , $r_{j}=0$ too. This shows that $w_{1},\\ldots,w_{d}$ are left linearly independent, which means that $d$ is preserved by additions of columns.

Preservation of other ranks of $M$ are proved similarly.∎

Proposition 3.

Left (right) non-zero row scalar multiplication preserves left (right) row rank of $M$ ; left (right) non-zero column scalar multiplications preserves left (right) column rank of $M$ .

Proof.

Let $w_{1},\\ldots,w_{n}$ be vectors obtained respectively from $v_{1},\\ldots,v_{n}$ such that the $i$ -th vector $w_{i}=rv_{i}$ , where $0\eq r\\in D$ , and all other $w_{j}$ ’s are the same as the $v_{j}$ ’s. Assume that the first $d$ rows of $M$ are left linearly independent, and that $i\\leq d$ . Suppose $r_{1}w_{1}+\\cdots+r_{d}w_{d}=0$ . Then $r_{1}v_{1}+\\cdots+r_{i}(rv_{i})+\\cdots r_{d}v_{d}=0$ , which implies $r_{1}=\\cdots=r_{i}r=\\cdots=r_{d}=0$ . Since $r\eq 0$ , $r_{i}=0$ , and therefore $w_{1},\\ldots,w_{d}$ are left linearly independent.

The others are proved similarly.∎

Proposition 4.

Left (right) non-zero row scalar multiplication preserves right (left) column rank of $M$ ; left (right) non-zero column scalar multiplication preserves right (left) row rank of $M$ .

Proof.

Let us prove that right multiplying a column by a non-zero scalar $r$ preserves the left row rank $d$ of $M$ . The others follow similarly.

Let $w_{1},\\ldots,w_{n}$ be vectors obtained respectively from $v_{1},\\ldots,v_{n}$ such that the $i$ -th coordinate $b_{ik}$ of $w_{k}$ is $a_{ik}r$ , where $a_{ik}$ is the $i$ -th coordinate of $v_{k}$ . Suppose once again that the first $d$ rows of $M$ are left linearly independent, and suppose $r_{1}w_{1}+\\cdots+r_{d}w_{d}=0$ . Then for each coordinate $j$ we get an equation $r_{1}b_{1j}+\\cdots+r_{d}b_{dj}=0$ . In particular, for the $i$ -th coordinate, we have $r_{1}a_{1j}r+\\cdots+r_{d}a_{dj}r=0$ . Since $r\eq 0$ , right multiplying the equation by $r^{-1}$ gives us $r_{1}a_{1j}+\\cdots+r_{d}a_{dj}=0$ . Re-collecting all the equations, we get $r_{1}v_{1}+\\cdots+r_{d}w_{d}=0$ , which implies that $r_{1}=\\cdots=r_{d}=0$ , or that $w_{1},\\ldots,w_{d}$ are left linearly independent.∎